Piezoelectric Sensors Change with Frequency

Discussion in 'General Electronics Chat' started by rperea, Oct 16, 2009.

  1. rperea

    Thread Starter Member

    Apr 7, 2009
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    I am trying to explain why the voltage increases when the frequency increases. I am testing a piezoplate at 1Hz and 2Hz, 10 cycles and a force of 30N. I collect the voltage difference. The votlage increases with frequency.

    The equations for piezoelectricity are:

    ε = SE * σ – da 33* Efield
    Electric Displacement = d33a * σ + (Permitivity) * Efield {Eq. 4}


    So.....what parameter depends on frequency? Is it the permitivity?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Probably the piezo element is rate sensitive. Shake it faster and get a higher output for a given force applied.

    There is probably some resonant point where the mechanical energy is coupled most efficiently - a simple experiment should should a peak output at that frequency.
     
    Last edited: Oct 16, 2009
  3. rperea

    Thread Starter Member

    Apr 7, 2009
    15
    0
    Thanks for the reply. What do you mean rate senstitive? I knoe the higher the frequency the higher the voltage I will get....now I am trying to understand this concept in the constitutive equations.....where thats the frequency play?
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    The higher the frequency, the higher the rate of change.
     
  5. BillB3857

    Senior Member

    Feb 28, 2009
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    Assuming you are using a uniform level of mechanical excitation, and plot the distance of excursion vs time, you will see that the rate of change in position is faster at higher frequencies.
     
  6. rperea

    Thread Starter Member

    Apr 7, 2009
    15
    0
    Again, is permitivity a function of frequency?
    Where does the frequency rate play the role in the consitutive equations?

    Sorry to be so confusing.
     
  7. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
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    I found this equivalent circuit of a piezo transducer. The series capacitance will explain the increased output at increased frequency. Xc=1/2πfc.
     
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